Chapter 11, Problem 76CR

a)

To determine

To evaluate: the slope of the graph of f at the given point.

Answer to Problem 76CR

Slope at (2, 14) is 4.

Explanation of Solution

Given:

  $f\left(x\right)=x^2+10,\left(2,14\right)$

Calculation: -

It is given that,

  $\begin{array}{l}f\left(x\right)=x^2+10\\ f\text{'}\left(x\right)=2x\\ f\text{'}\left(2\right)=4\end{array}$

Hence, Slope at (2, 14) is 4.

b)

To determine

To evaluate: the equation of the tangent line to the graph at the point.

Answer to Problem 76CR

The equation of tangent is $4x-y+6=0$ .

Explanation of Solution

Given:

  $f\left(x\right)=2x^2-1,\left(3,17\right)$

Calculation: -

It is given that,

  $\begin{array}{l}f\left(x\right)=x^2+10\\ f\text{'}\left(x\right)=2x\\ f\text{'}\left(2\right)=4\end{array}$

Hence, Slope at (2, 14) is 4.

Using one-point formula to find the equation of line,

  $\begin{array}{c}y-y_1=m\left(x-x_1\right)\\ y-14=4\left(x-2\right)\\ y-14=4x-8\\ 4x-y+6=0\end{array}$

Hence, the equation of tangent is $4x-y+6=0$ .

c)

To determine

To evaluate: the graph of the function and the tangent line on it.

Answer to Problem 76CR

  

Explanation of Solution

Given:

  $f\left(x\right)=2x^2-1,\left(3,17\right)$

Calculation: -

It is given that,

  $\begin{array}{l}f\left(x\right)=2x^2-1\\ f\text{'}\left(x\right)=4x\\ f\text{'}\left(3\right)=12\end{array}$

Hence, Slope at (3, 17) is 12.

Using one-point formula to find the equation of line,

  $\begin{array}{c}y-y_1=m\left(x-x_1\right)\\ y-17=12\left(x-3\right)\\ y-17=12x-36\\ 12x-y-19=0\end{array}$

Hence, the equation of tangent is $12x-y-19=0$ .

Using graphing calculator to plot the graph,